What is the mass of the planet?

  • Thread starter Unknown221
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  • #1
I have a few questions that I'm not sure on how to do:

1) A 20kg mass is connected to a 8kg mass by a thin frictionless string on a pulley. It accelerates to 1.5m/s2 and the heavier mass slides along a rough surface. What is the coefficient of friction?

3) 1.00g of U-238 fuel is consumed. Assuming 100% efficiency, what is the nuclear energy used in joules?

4) A 7.5x10^3kg sattelite orbits a planet at an altitude of 300km. The radius of the planet is 8.5x10^6. The gravitational field at the radius is 6.5N/kg. What is the mass of the planet?

5)A rocket car uniformly accelterates at 15m/s2 and heads towards a cliff that is 75m high. At what distance should the landing pad be placed so the car can land?

Need help with these questions ASAP.. Thanks.
 

Answers and Replies

  • #2
I am so tired of people posting threads like this. You have obviously not read the guidelines regarding the posting of homework questions.

I have a few questions that I'm not sure on how to do:

1) A 20kg mass is connected to a 8kg mass by a thin frictionless string on a pulley. It accelerates to 1.5m/s2 and the heavier mass slides along a rough surface. What is the coefficient of friction?

3) 1.00g of U-238 fuel is consumed. Assuming 100% efficiency, what is the nuclear energy used in joules?

4) A 7.5x10^3kg sattelite orbits a planet at an altitude of 300km. The radius of the planet is 8.5x10^6. The gravitational field at the radius is 6.5N/kg. What is the mass of the planet?

5)A rocket car uniformly accelterates at 15m/s2 and heads towards a cliff that is 75m high. At what distance should the landing pad be placed so the car can land?

Need help with these questions ASAP.. Thanks.
 
  • #3
I am so tired of people posting threads like this. You have obviously not read the guidelines regarding the posting of homework questions.

Sorry I have a test tomorrow and am lost on how to do these questions.. Just need some help on how to get started..
 
  • #4
I will certainly try to help you, but you have to show me that you've at least thought about these questions prior to asking for help. We're here to help you learn, not do your homework for you.

Sorry I have a test tomorrow and am lost on how to do these questions.. Just need some help on how to get started..
 
  • #5
2) 3) 1.00g of U-238 fuel is consumed. Assuming 100% efficiency, what is the nuclear energy used in joules?
For this one would I use E=mc^2 and just fill it in with the data provided?

5) A rocket car uniformly accelerates at 15m/s2 and heads towards a cliff that is 75m high. At what distance should the landing pad be placed so the car can land?

For this one I think I have the answer. We know:
X:
a= 15m/s2
d=?
vi= 0m/s

Y:
d= -75m
a= -9.8m/s2
vi= 0.0m/s
t=?

Y: d= vit+.5at^2
and when solving for t, I got 2.8s.

then for X: d= vit+.5at^2
a= 15m/s2
d=?
vi= 0m/s
t= 2.8s

Answer: d= 59m

As for the other two, I do not know what to do for them as #1 I cannot find examples for and #4 I haven't done before so I'm completely lost on those and these are the ones that will be on the exam, so I just need something to get me started on them, thanks.
 
  • #6
Anyone able to help me out?
 
  • #7
nrqed
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2) 3) 1.00g of U-238 fuel is consumed. Assuming 100% efficiency, what is the nuclear energy used in joules?
For this one would I use E=mc^2 and just fill it in with the data provided?
Yes. Make sure you put the mass in kilograms
5) A rocket car uniformly accelerates at 15m/s2 and heads towards a cliff that is 75m high. At what distance should the landing pad be placed so the car can land?
This question does not make sense to me. If we don't know for how long it accelerates before reaching the cliff, we don't know its horizontal speed when it goes over the cliff and it is impossible to find how far it will land.
 
  • #8
Yes. Make sure you put the mass in kilograms

This question does not make sense to me. If we don't know for how long it accelerates before reaching the cliff, we don't know its horizontal speed when it goes over the cliff and it is impossible to find how far it will land.

*The distance from the car to the edge of the cliff is 100m.
 
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  • #9
nrqed
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*The distance from the car to the edge of the cliff is 100m.

:eek: Ah!! It was not in the original question.

Ok, but then your solution is wrong. You did not use hat information. And what you did is to use the time of fall in the vertical direction (2.8 sec) as the time for the horziontal motion. There is no reason to assume that those two times are equal!

What you need to do:

a) find the horizontal velocity when the car reaches the cliff using 100 m and the 15 m/s^2 (are you sure it's 15!?! This is too large for a real car)

b) find the time it takes to fall 75 m. That you already did. I did not check but if you did it correctly, it will be 2.8 s (you might want to keep more sig figs)

c) finally, use the x velcoity of part a and the time of part b to find the horizontal distance travelled during the fall
 
  • #10
nrqed
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:eek: Ah!! It was not in the original question.

Ok, but then your solution is wrong. You did not use hat information. And what you did is to use the time of fall in the vertical direction (2.8 sec) as the time for the horziontal motion. There is no reason to assume that those two times are equal!

What you need to do:

a) find the horizontal velocity when the car reaches the cliff using 100 m and the 15 m/s^2 (are you sure it's 15!?! This is too large for a real car)

b) find the time it takes to fall 75 m. That you already did. I did not check but if you did it correctly, it will be 2.8 s (you might want to keep more sig figs)

c) finally, use the x velcoity of part a and the time of part b to find the horizontal distance travelled during the fall

Unknown221 said:
Hey I was wondering if you could show me a few steps for the problem I asked because I looked through my notes and we never learned how to find horizontal distances and the velocity was always given to us..

X:
d= 100m
a= 15m/s^2 (you missed that its a rocket car)
v=?
t=?

Y:
d=-75m
a=-9.8m/s^2
vi=0.0
t=2.8

could you please show me to find the answer, thanks.


The equations are

Along x

Eq 1: [itex] x_f = x_i + v_{xi} t + 1/2 a_x t^2 [/itex]

Eq 2: [itex] v_{xf} = v_{xi} + a_x t [/itex]


From the previous two, one can also show that

Eq 3: [itex] v_{xf}^2 = v_{xi}^2 + 2 a_x (x_f-x_i) [/itex]

Along y (assuming only the force of gravity is present)

Eq 4 [itex] y_y = y_i + v_{yi} t - 1/2 g t^2 [/itex]

Eq 5 [itex] v_{yf} = v_{yi} - g t [/itex]

Eq 3: [itex] v_{yf}^2 = v_{yi}^2 - 2 g (y_f-y_i) [/itex]
where I use g = +9.8 m/s^2 (my g is positive).


Along x, if you now that the distance (x_f -x_i) is 100 m and you know a_x, and v_xi=0, use eq 4 to find the value of v_xf when the car reaches the cliff



Hope this helps
 
  • #11
I got 54.8m/s as the velocity and before you said that the x and y time are not the same so should I use the y time to solve for the distance? Using the y time I get 157.4 as the ditance. Nvm got diff answer and gotta go now, thanks for trying to help me though.
 
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